Subtracting the first sum on the right from both sides and using the equality of mixed partials,

( )
∑ g ( ∑ (cof (Dg)) ) = 0.
i i,k j ij,j

If det

(gi,k)

≠0 so that

(gi,k)

is invertible, this shows ∑j

(cof (Dg ))

ij,j = 0. If det

(Dg)

= 0,
let

g (x) = g (x) + εx
k k

where εk→ 0 and det

(Dg + εkI)

≡ det

(Dgk )

≠0. Then

∑ ∑
(cof (Dg ))ij,j = lim (cof (Dgk ))ij,j = 0 ■
j k→∞ j

Another simple lemma which will be used whenever convenient is the following lemma.

Lemma 11.2.2Let K be a compact set and C a closed set in ℝpsuch that K ∩ C = ∅. Then

dist(K,C ) ≡ inf{∥k− c∥ : k ∈ K, c ∈ C } > 0.

Proof:Let

d ≡ inf {||k− c|| : k ∈ K, c ∈ C}

Let

{ki}

,

{ci}

be such that

1
d + i > ||ki − ci||.

Since K is compact, there is a subsequence still denoted by

{k }
i

such that ki→ k ∈ K. Then
also

||ci − cm|| ≤ ||ci − ki||+ ||ki − km ||+ ||cm − km ||

If d = 0, then as m,i →∞ it follows

||c − c ||
i m

→ 0 and so

{c }
i

is a Cauchy sequence which must
converge to some c ∈ C. But then

||c − k||

= limi→∞

||c − k ||
i i

= 0 and so c = k ∈ C ∩K, a contradiction
to these sets being disjoint. ■

In particular the distance between a point and a closed set is always positive if the point is not in the
closed set. Of course this is obvious even without the above lemma. The above lemmas will be used now to
prove technical lemmas which are the basis for everything.

Definition 11.2.3Let g ∈ C∞

(-- )
Ω;ℝp

where Ω is a bounded open set. Also let ϕεbe a mollifier.

∫
ϕε ∈ C ∞c (B(0,ε)),ϕε ≥ 0, ϕεdx = 1.

First, here is a technical lemma which will end up being the reason one can define the
degree in terms of an integral. It is a result on homotopy invariance for functions which are
C∞.

Lemma 11.2.4If h : ℝp×ℝ → ℝpis in Cc∞

(ℝp × ℝ,ℝp)

, and0

∕∈

h

(∂Ω× [α,β])

then for
0 < ε <dist

(0,h(∂Ω × [α,β ]))

,

∫
t → Ω ϕε(h(x,t))detD1h (x,t)dx

is constant for t ∈

(a,b)

, an open set which contains

[α,β]

.

Proof:By continuity of h, h

(∂Ω × [α,β ])

is compact and so is at a positive distance from 0.
Therefore, there exists an open interval,

(a,b)

such that

(a,b)

contains

[α,β]

and 0

∕∈

h

(∂Ω × [a,b])

. Let
ε > 0 be such that for all t ∈

[a,b]

,

B (0,ε)∩ h(∂Ω × [a,b]) = ∅ (11.5)

(11.5)

Define for t ∈

(a,b)

,

∫
H (t) ≡ Ωϕε(h (x,t))detD1h (x,t)dx

Then if t ∈

(a,b)

,

∫ ∑
H ′(t) = ϕε,α (h(x,t))hα,t(x,t)detD1h (x,t)dx
Ω α

∫ ∑
+ ϕε(h (x,t)) detD1 (h (x,t)),αjh α,jtdx
Ω α,j

≡ A + B.

In this formula, the function det is considered as a function of the n2 entries in the n × n matrix and the
,αj represents the derivative with respect to the αjth entry hα,j. Now as in the proof of Lemma 11.2.1 on
Page 873,

Now the sum on j is the dot product of the βth row with the αth row of the cofactor matrix
which equals zero unless β = α because it would be a cofactor expansion of a matrix with
two equal rows. When β = α, the sum on j reduces to det